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vladimir2022 [97]
4 years ago
12

Quadrilateral ABCD has vertices A(-2,4), B(-3.-3), C(-5,-2), D(-5,2). Which of the following are the coordinates of A’ after a d

ilation centered at the origin with a scale factor of 2 1/2.

Mathematics
1 answer:
Gala2k [10]4 years ago
3 0
<h2>Answer:</h2>

\boxed{A'(-5,10)}

<h2>Step-by-step explanation:</h2>

Dilation is when you stretch something by the same amount in two perpendicular directions. In other words, dilation changes size, not overall shape. Dilation factors greater than one makes the shape greater while dilation factors less than one makes the shape smaller. The scale factor in this problem is:

2\frac{1}{2}

Which is a mixed fraction. This can be transformed into an improper fraction as follows:

2\frac{1}{2}=2+\frac{1}{2}=\frac{4+1}{2}=\frac{5}{2}

So the scale factor is \frac{5}{2}. Since the center of dilation is the origin, we just need to multiply each vertex of the quadrilateral by the scale factor. Thus, for A we have:

A'(-2\times \frac{5}{2},4\times \frac{5}{2})=\boxed{A'(-5,10)}

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Volgvan

Answer:

\displaystyle y=\frac{16-9x^3}{2x^3 - 3}

\displaystyle y=-\frac{56}{13}

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}

i)

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3

Simplify the radical with the cube:

\displaystyle x^3=\frac{3y+16}{2y+9}

Multiply by 2y+9

\displaystyle x^3(2y+9)=\frac{3y+16}{2y+9}(2y+9)

Simplify:

\displaystyle x^3(2y+9)=3y+16

Operate the parentheses:

\displaystyle x^3(2y)+x^3(9)=3y+16

\displaystyle 2x^3y+9x^3=3y+16

Subtract 3y and 9x^3:

\displaystyle 2x^3y - 3y=16-9x^3

Factor y out of the left side:

\displaystyle y(2x^3 - 3)=16-9x^3

Divide by 2x^3 - 3:

\mathbf{\displaystyle y=\frac{16-9x^3}{2x^3 - 3}}

ii) To find y when x=2, substitute:

\displaystyle y=\frac{16-9\cdot 2^3}{2\cdot 2^3 - 3}

\displaystyle y=\frac{16-9\cdot 8}{2\cdot 8 - 3}

\displaystyle y=\frac{16-72}{16- 3}

\displaystyle y=\frac{-56}{13}

\mathbf{\displaystyle y=-\frac{56}{13}}

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3 years ago
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Step-by-step explanation:

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